2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constantsK, = 3 and K, = 5, as shown in Figure. Parallel-plates have the area A =d = 0.05 m. A potential difference of AV = 24 V is applied to the circuit.(a) Find the equivalent capacitance of the system.(b) Find the potential AV1. (E,-8.85×10-12 C²/N×m²)0.7 m². The distance isK1A/2AV1K2A/2Аd.K2 | K1K1A/22d/3 d/3K2A/2dAV = 24 V

This question is based on combination of capacitors.

2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants Kj = 3 and K2 = 5, as shown in Figure. Parallel-plates have the area A = 0.7 m². The distance is d= 0.05 m. A potential difference of AV (a) Find the equivalent capacitance of the system. (b) Find the potential AVj. (E-8.85×10-12 C²/N×m²) = 24 V is applied to the circuit. K1 A/2 AV1 K2 A/2

2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants Kj = 3 and K2 = 5, as shown in Figure. Parallel-plates have the area A = 0.7 m². The distance is d= 0.05 m. A potential difference of AV (a) Find the equivalent capacitance of the system. (b) Find the potential AVj. (E-8.85×10-12 C²/N×m²) = 24 V is applied to the circuit. K1 A/2 AV1 K2 A/2

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter20: Electric Potential And Capacitance

Section: Chapter Questions

Problem 81P

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Similar questions

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